1. Lecture 15 Two-Factor Analysis of Variance (Chapter 15.5)

2. Oneway Analysis of Sales By Strategy Comparisons for all pairs using Tukey-Kramer HSD Abs(Dif)-LSD QualityPriceConv. Quality -71.768 -27.418 3.682 Price -27.418 -71.76 -40.668 Conven. 3.682 -40.668 -71.768 Positive values show pairs of means that are significantly different.

3. BlocksTreatmentsb-1MST / MSEMSB / MSE Conclusion: At 5% significance level there is sufficient evidence to infer that the mean “cholesterol reduction” gained by at least two drugs are different. K-1 Randomized Blocks ANOVA - Example

4. 15.5 Two-Factor Analysis of Variance - Example 15.3 –Suppose in Example 15.1, two factors are to be examined: The effects of the marketing strategy on sales. –Emphasis on convenience –Emphasis on quality –Emphasis on price The effects of the selected media on sales. –Advertise on TV –Advertise in newspapers

5. Solution –We may attempt to analyze combinations of levels, one from each factor using one-way ANOVA. –The treatments will be: Treatment 1: Emphasize convenience and advertise in TV Treatment 2: Emphasize convenience and advertise in newspapers ……………………………………………………………………. Treatment 6: Emphasize price and advertise in newspapers Attempting one-way ANOVA

6. Solution –The hypotheses tested are: H 0 :  1 =  2 =  3 =  4 =  5 =  6 H 1 : At least two means differ. Attempting one-way ANOVA

7. City1 City2 City3 City4City5City6 Convnce Convnce Quality Quality Price Price TVPaper TV Paper TV Paper – In each one of six cities sales are recorded for ten weeks. – In each city a different combination of marketing emphasis and media usage is employed. Solution Attempting one-way ANOVA

8. The p-value =.0452. We conclude that there is evidence that differences exist in the mean weekly sales among the six cities. City1 City2 City3 City4City5City6 Convnce Convnce Quality Quality Price Price TVPaper TV Paper TV Paper Solution Xm15-03 Attempting one-way ANOVA

9. These result raises some questions: –Are the differences in sales caused by the different marketing strategies? –Are the differences in sales caused by the different media used for advertising? –Are there combinations of marketing strategy and media that interact to affect the weekly sales? Interesting questions – no answers

10. The current experimental design cannot provide answers to these questions. A new experimental design is needed. Two-way ANOVA (two factors)

11. City 1 sales City3 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers ConvenienceQualityPrice Are there differences in the mean sales caused by different marketing strategies? Factor A: Marketing strategy Factor B: Advertising media

12. Test whether mean sales of “Convenience”, “Quality”, and “Price” significantly differ from one another. H 0 :  Conv. =  Quality =  Price H 1 : At least two means differ Calculations are based on the sum of square for factor A SS(A) Two-way ANOVA (two factors)

13. City 1 sales City 3 sales City 5 sales City 2 sales City 4 sales City 6 sales Factor A: Marketing strategy Factor B: Advertising media Are there differences in the mean sales caused by different advertising media? TV Newspapers ConvenienceQualityPrice

14. Test whether mean sales of the “TV”, and “Newspapers” significantly differ from one another. H 0 :  TV =  Newspapers H 1 : The means differ Calculations are based on the sum of square for factor B SS(B) Two-way ANOVA (two factors)

15. City 1 sales City 5 sales City 2 sales City 4 sales City 6 sales TV Newspapers ConvenienceQualityPrice Factor A: Marketing strategy Factor B: Advertising media Are there differences in the mean sales caused by interaction between marketing strategy and advertising medium? City 3 sales TV Quality

16. Test whether mean sales of certain cells are different than the level expected. Calculation are based on the sum of square for interaction SS(AB) Two-way ANOVA (two factors)

17. Levels of factor A 123 Level 1 of factor B Level 2 of factor B 123 123123 Level 1and 2 of factor B Difference between the levels of factor A No difference between the levels of factor B Difference between the levels of factor A, and difference between the levels of factor B; no interaction Levels of factor A No difference between the levels of factor A. Difference between the levels of factor B Interaction M R e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e M R e s a p n o n s e

18. Sums of squares

19. F tests for the Two-way ANOVA Test for the difference between the levels of the main factors A and B F= MS(A) MSE F= MS(B) MSE Rejection region: F > F ,a-1,n-ab F > F , b-1, n-ab Test for interaction between factors A and B F= MS(AB) MSE Rejection region: F > F  a-1)(b-1),n-ab SS(A)/(a-1) SS(B)/(b-1) SS(AB)/(a-1)(b-1) SSE/(n-ab)

20. Required conditions: 1.The response distributions is normal 2.The treatment variances are equal. 3.The samples are independent random samples.

21. Example 15.3 – continued( Xm15-03) Xm15-03 F tests for the Two-way ANOVA

22. Example 15.3 – continued –Test of the difference in mean sales between the three marketing strategies H 0 :  conv. =  quality =  price H 1 : At least two mean sales are different F tests for the Two-way ANOVA Factor A Marketing strategies

23. Example 15.3 – continued –Test of the difference in mean sales between the three marketing strategies H 0 :  conv. =  quality =  price H 1 : At least two mean sales are different F = MS(Marketing strategy)/MSE = 5.33 F critical = F ,a-1,n-ab = F.05,3-1,60-(3)(2) = 3.17; (p-value =.0077) –At 5% significance level there is evidence to infer that differences in weekly sales exist among the marketing strategies. F tests for the Two-way ANOVA MS(A)  MSE

24. Example 15.3 - continued –Test of the difference in mean sales between the two advertising media H 0 :  TV. =  Nespaper H 1 : The two mean sales differ F tests for the Two-way ANOVA Factor B = Advertising media

25. Example 15.3 - continued –Test of the difference in mean sales between the two advertising media H 0 :  TV. =  Nespaper H 1 : The two mean sales differ F = MS(Media)/MSE = 1.42 F critical = F  a-1,n-ab = F.05,2-1,60-(3)(2) = 4.02 (p-value =.2387) –At 5% significance level there is insufficient evidence to infer that differences in weekly sales exist between the two advertising media. F tests for the Two-way ANOVA MS(B)  MSE

26. Example 15.3 - continued –Test for interaction between factors A and B H 0 :  TV*conv. =  TV*quality =…=  newsp.*price H 1 : At least two means differ F tests for the Two-way ANOVA Interaction AB = Marketing*Media

27. Example 15.3 - continued –Test for interaction between factor A and B H 0 :  TV*conv. =  TV*quality =…=  newsp.*price H 1 : At least two means differ F = MS(Marketing*Media)/MSE =.09 F critical = F  a-1)(b-1),n-ab = F.05,(3-1)(2-1),60-(3)(2) = 3.17 (p-value=.9171) –At 5% significance level there is insufficient evidence to infer that the two factors interact to affect the mean weekly sales. MS(AB)  MSE F tests for the Two-way ANOVA